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Anon25 [30]
3 years ago
13

Alchemists in the Middle Ages dreamed of converting base metals, such as lead, into precious metals—gold and silver. Why could t

hey never succeed? Today could we convert lead into gold?
Chemistry
1 answer:
lesantik [10]3 years ago
4 0

Explanation:

Each element in the periodic table has different but fixed number of the protons in nucleus of it's atom, which is known as the atomic number.

Transmutation of one chemical element into the another involves the changing of the atomic number. Such nuclear reaction requires millions of the times more energy as compared to normal chemical reactions. Thus, the dream of  the alchemist of transmuting the lead into the gold was never achievable chemically .

Conversion of lead to gold in today's world:

This conversion is indeed possible. The requirements are a particle accelerator, tremendous supply of the energy. Nuclear scientists at the Lawrence Berkeley National Laboratory located in California, more than 30 years ago, succeeded in producing very minute amounts of the gold from the bismuth. Bismuth is a metallic element which is adjacent to the lead on periodic table. Same process would work for the lead but isolating gold at end of reaction would prove much more difficult because lead is available in many isotopes. The homogeneous nature of the element means that it is easier to separate the gold from the bismuth as compared to separate the gold from the lead which has four  isotopic identities which all are stable.

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How many grams are in 11.9 moles of chromium?
raketka [301]

Explanation:

It is known that one mole of chromium or molar mass of chromium is 51.99 g/mol.

It is given that number of moles is 11.9 moles.

Therefore, calculate the mass of chromium in grams as follows.

     No. of moles = \frac{mass in grams}{Molar mass}

    mass in grams = No. of moles × Molar mass

                             = 11.9 moles × 51.99 g/mol

                             = 618.68 g

Thus, we can conclude that there are 618.68 g in 11.9 moles of chromium.

7 0
3 years ago
Read 2 more answers
A cup of coffee has 71 mL of coffee and 127 mL of water. What is the percent volume of the coffee solution?
Anna [14]

Answer:

35.9%

Explanation:

The percent volume of the coffee solution can be calculated as follows:

% volume of coffee solution = volume of coffee/total volume of coffee solution × 100

According to this question, a cup of coffee has 71 mL of coffee and 127 mL of water. This means that, the total volume of coffee solution is;

71mL + 127mL = 198mL

% volume = 71/198 × 100

= 0.359 × 100

Percent volume of coffee solution = 35.9%

3 0
3 years ago
The specific heat of liquid ethyl alcohol is 2.42 j/g C and its density of 0.7893 g/mL. A piece of solid sliver (specific heat 0
Doss [256]
MAg*cAg*(T1-T)=ρalc*Valc*calc*(T-T2)
mAg=?(g)
cAg=0.24J/gC
T1=95
T=23.5
Valc=25.6ml
ρalc=0.7893g/ml
T2=19.27
use wollframalpha or calculator
6 0
2 years ago
Read 2 more answers
Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine
Lilit [14]

<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 75.77% and 24.23% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{17}^{35}\textrm{Cl} isotope be 'x'. So, fractional abundance of _{17}^{37}\textrm{Cl} isotope will be '1 - x'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

Mass of _{17}^{35}\textrm{Cl} isotope = 34.9689 amu

Fractional abundance of _{17}^{35}\textrm{Cl} isotope = x

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

Mass of _{17}^{37}\textrm{Cl} isotope = 36.9659 amu

Fractional abundance of _{17}^{37}\textrm{Cl} isotope = 1 - x

  • Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577

Percentage abundance of _{17}^{35}\textrm{Cl} isotope = 0.7577\times 100=75.77\%

Percentage abundance of _{17}^{37}\textrm{Cl} isotope = (1-0.7577)=0.2423\times 100=24.23\%

Hence, the percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 75.77% and 24.23% respectively.

6 0
2 years ago
The word alkyl is short for
Colt1911 [192]

Answer:

An alkyl is a functional group of an organic chemical that contains only carbon and hydrogen atoms, which are arranged in a chain.

Explanation:

4 0
2 years ago
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